Find a number such that number of divider is $16$ and sum of divider is $4032$

Question

I have solution where it write that we only can have a number in this shape $pqrs,p^3qr,p^3q^3,p^7q,p^{15}$. I understand why we can have only four divider but why we can have only number in that shape, why I can not write for example $p^2q^5$ and I still have $16$ divider, or $p^2qr^2$ and I still have $16$ divider?

Answer 1

$p^2q^5$ has $(2+1)(5+1)=18$ divisors $$1,q,q^2,q^3,q^4,q^5,p,pq,pq^2,pq^3,pq^4,pq^5,p^2,p^2q,p^2q^2,p^2q^3,p^2q^4,p^2q^5.$$ In other words, you want that the product of the incremented-by-one exponents is $16$. You only have to consider the factorizations $2\cdot 2\cdot 2\cdot 2=4\cdot 2\cdot 2=4\cdot 4=8\cdot 2=16$, corresponding exactly to the prime patterns listed.